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Mathematics

If the mean of a and 1a\dfrac{1}{a} is x, then the mean of a3 and 1a3\dfrac{1}{a^3} is :

  1. x(4x2 - 3)

  2. x(x2- 2)

  3. x2 + 3

  4. x2

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Answer

Mean of a and 1a\dfrac{1}{a} is x

⇒a+1a2=x\Rightarrow \dfrac{a + \dfrac{1}{a}}{2} = x

⇒a+1a=2x\Rightarrow a + \dfrac{1}{a} = 2x

We know that,

(a + b)3 = a3 + b3 + 3ab(a + b)

Substituting b=1ab = \dfrac{1}{a} :

⇒(a+1a)3=a3+1a3+3(a)(1a)(a+1a)⇒(a+1a)3=a3+1a3+3(a+1a)\Rightarrow \left(a + \dfrac{1}{a}\right)^3 = a^3 + \dfrac{1}{a^3} + 3(a)\left(\dfrac{1}{a}\right)\left(a + \dfrac{1}{a}\right) \\[1em] \Rightarrow \left(a + \dfrac{1}{a}\right)^3 = a^3 + \dfrac{1}{a^3} + 3\left(a + \dfrac{1}{a}\right) \\[1em]

Substituting a+1a=2xa + \dfrac{1}{a} = 2x into the identity:

⇒(2x)3=a3+1a3+3(2x)⇒8x3=a3+1a3+6x⇒a3+1a3=8x3−6x\Rightarrow (2x)^3 = a^3 + \dfrac{1}{a^3} + 3(2x) \\[1em] \Rightarrow 8x^3 = a^3 + \dfrac{1}{a^3} + 6x \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = 8x^3 - 6x \\[1em]

The mean of a3 and 1a3\dfrac{1}{a^3} is:

⇒Mean=a3+1a32⇒Mean=8x3−6x2⇒Mean=4x3−3x⇒Mean=x(4x2−3)\Rightarrow \text{Mean} = \dfrac{a^3 + \dfrac{1}{a^3}}{2} \\[1em] \Rightarrow \text{Mean} = \dfrac{8x^3 - 6x}{2} \\[1em] \Rightarrow \text{Mean} = 4x^3 - 3x \\[1em] \Rightarrow \text{Mean} = x(4x^2 - 3)

Hence, option 1 is the correct option.

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